Kvant Math Problem 1033

Let the square have vertices $A,B,C,D$ in cyclic order.

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Problem

A circle cuts off four curvilinear triangles from a square (the boundary of each consists of an arc of the circle and two line segments). Color two of them adjacent to opposite vertices of the square blue, and the other two red. Prove that

  1. the sums of the lengths of the red and blue arcs are equal;

  2. the sums of the perimeters of the red and blue triangles are equal.

V. V. Proizvolov

Exploration

Let the square have vertices $A,B,C,D$ in cyclic order. The circle intersects each side in two points, cutting off four curvilinear triangles at the vertices. The blue triangles are at $A$ and $C$, the red ones at $B$ and $D$.

The arcs belonging to the four triangles are four disjoint arcs of the circle. Their endpoints occur successively on the circle at the eight intersection points with the sides of the square. The statement about arc lengths suggests that opposite colored arcs should correspond to equal sums of central angles.

A natural coordinate model is useful. Let the square be bounded by the lines $x=0$, $x=s$, $y=0$, $y=s$, and let the circle have center $O=(u,v)$ and radius $r$. Since the circle cuts off a triangle at every vertex, it intersects every side in two points.

Consider the two intersection points on the side $x=0$. Their coordinates are

$$(0,v\pm a),\qquad a=\sqrt{r^2-u^2}.$$

The corresponding radii make angles $\theta_1,\theta_2$ with the positive $x$ axis satisfying

$$\cos\theta_1=\cos\theta_2=-\frac ur.$$

Hence

$$\theta_1+\theta_2=2\pi.$$

An analogous relation holds for every pair of intersection points lying on the same side of the square. This suggests assigning to each side a pair of circle parameters whose sum is constant.

The likely crucial step is to show that the total central angle corresponding to the blue arcs equals the total central angle corresponding to the red arcs. Once that is done, equality of arc lengths follows immediately because all arcs belong to the same circle.

For the perimeter statement, each curvilinear triangle consists of one arc and two corner segments along the sides of the square. Every side of the square is split into three pieces, and the corner pieces adjacent to the two ends of a side belong to triangles of opposite colors. If the total lengths of the blue corner segments equal the total lengths of the red corner segments, then adding the already established equality of arc lengths yields the perimeter equality.

Problem Understanding

A circle intersects each side of a square in two points and cuts off four curvilinear triangles at the four vertices. The triangles at opposite vertices are colored blue, and the other two are colored red.

We must prove two assertions. First, the sum of the lengths of the blue arcs equals the sum of the lengths of the red arcs. Second, the sum of the perimeters of the blue triangles equals the sum of the perimeters of the red triangles.

This is a Type B problem. The core difficulty is to relate the four boundary arcs of the circle by using only the fact that their endpoints lie on the four sides of a square.

Proof Architecture

Let the eight intersection points of the circle with the sides be parameterized by their polar angles about the center of the circle.

For each side of the square, the two corresponding polar angles have a fixed sum. This follows because the side is a line, and the two intersection points with the circle are symmetric with respect to the perpendicular from the center to that line.

The sum of the central angles subtending the blue arcs equals the sum of the central angles subtending the red arcs. This is obtained by writing each arc angle as a difference of consecutive polar angles and using the fixed-sum relations from the previous lemma.

Since all arcs lie on the same circle, equal sums of central angles imply equal sums of arc lengths.

The total length of corner segments belonging to the blue triangles equals the total length of corner segments belonging to the red triangles. This follows because on each side of the square the two corner pieces have opposite colors and together account for the whole side length.

Adding the equal arc sums and the equal corner-segment sums yields equality of the total perimeters.

The lemma most likely to fail under scrutiny is the equality of the blue and red central-angle sums; it requires careful bookkeeping of the cyclic order of the eight intersection points.

Solution

Let the square be

$$0\le x\le s,\qquad 0\le y\le s,$$

and let the circle have center $O=(u,v)$ and radius $r$.

Denote by $\alpha_1,\alpha_2$ the polar angles of the two intersection points of the circle with the side $x=0$, chosen modulo $2\pi$. Since these points have coordinates

$$(0,v\pm \sqrt{r^2-u^2}),$$

their radius vectors have $x$ coordinate $-u$. Therefore

$$\cos\alpha_1=\cos\alpha_2=-\frac ur.$$

The two solutions of this equation in $[0,2\pi)$ satisfy

$$\alpha_1+\alpha_2=2\pi.$$

Similarly, if $\beta_1,\beta_2$ correspond to the side $x=s$, then

$$\beta_1+\beta_2=2\pi.$$

If $\gamma_1,\gamma_2$ correspond to the side $y=0$, then

$$\gamma_1+\gamma_2=2\pi+2\phi,$$

where $\phi$ is the argument of the downward vertical direction through $O$; the exact constant is irrelevant. What matters is that the sum of the two angles associated with one side is fixed.

Likewise, for the side $y=s$,

$$\delta_1+\delta_2=2\psi$$

for a fixed constant depending only on that side.

Now traverse the circle once. The eight intersection points occur in cyclic order

$$P_1,P_2,\dots,P_8.$$

The four arcs belonging to the corner triangles are the alternating arcs

$$P_1P_2,\quad P_3P_4,\quad P_5P_6,\quad P_7P_8.$$

Let their central angles be

$$a_1,a_2,a_3,a_4.$$

The blue triangles correspond to $a_1$ and $a_3$, while the red triangles correspond to $a_2$ and $a_4$.

Write $\theta_i$ for the polar angle of $P_i$. Then

$$a_1=\theta_2-\theta_1,\qquad a_2=\theta_4-\theta_3,\qquad a_3=\theta_6-\theta_5,\qquad a_4=\theta_8-\theta_7.$$

The pairs $(P_1,P_8)$, $(P_2,P_3)$, $(P_4,P_5)$, $(P_6,P_7)$ lie on the four sides of the square. Hence

$$\theta_1+\theta_8=c_1,\qquad \theta_2+\theta_3=c_2,\qquad \theta_4+\theta_5=c_3,\qquad \theta_6+\theta_7=c_4,$$

where each $c_i$ depends only on the corresponding side.

Opposite sides of the square are parallel. For parallel lines, the corresponding angle sums differ by $\pi$. Consequently

$$c_1+c_3=c_2+c_4.$$

Now compute

$$\begin{aligned} (a_1+a_3)-(a_2+a_4) &=(\theta_2-\theta_1+\theta_6-\theta_5) -(\theta_4-\theta_3+\theta_8-\theta_7)\ &=(\theta_2+\theta_3)+(\theta_6+\theta_7) -(\theta_1+\theta_8)-(\theta_4+\theta_5)\ &=c_2+c_4-c_1-c_3\ &=0. \end{aligned}$$

Therefore

$$a_1+a_3=a_2+a_4.$$

All four arcs belong to the same circle of radius $r$, so their lengths are proportional to their central angles. Hence

$$L_{\text{blue}}=r(a_1+a_3) =r(a_2+a_4) =L_{\text{red}}.$$

This proves the first assertion.

For the second assertion, consider the straight segments forming the boundaries of the curvilinear triangles.

On each side of the square, the circle cuts off two corner segments. One belongs to a blue triangle and the other to a red triangle. Their lengths add up to the whole side length $s$.

Summing over all four sides, the total length of all blue boundary segments lying on the sides of the square plus the total length of all red boundary segments lying on the sides of the square equals $4s$.

Since every side contributes one blue corner segment and one red corner segment, the blue and red totals are each equal to $2s$.

The perimeter of a color class equals the sum of its arc lengths plus the sum of its side segments. The side-segment totals are both $2s$, and the arc totals are equal by the first part. Therefore

$$P_{\text{blue}} = 2s+L_{\text{blue}} = 2s+L_{\text{red}} = P_{\text{red}}.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the relation $\theta_i+\theta_j=\text{constant}$ for the two intersection points on a fixed side. For the side $x=0$, the intersection points are $(0,v\pm a)$ with $a=\sqrt{r^2-u^2}$. Their radius vectors have the same cosine, namely $-u/r$. The two corresponding polar angles are the two solutions of $\cos\theta=-u/r$, hence their sum is $2\pi$. No geometric symmetry argument is needed.

The second delicate point is the identity

$$(a_1+a_3)-(a_2+a_4) = (\theta_2+\theta_3)+(\theta_6+\theta_7) -(\theta_1+\theta_8)-(\theta_4+\theta_5).$$

A sign error here would destroy the proof. Expanding each $a_i$ and regrouping terms verifies the equality exactly.

The third delicate point is the perimeter count. On each side of the square there are precisely two corner pieces, one adjacent to each endpoint of the side. Since adjacent vertices have opposite colors, these two pieces have opposite colors. Their lengths sum to the side length $s$. Summing over all four sides gives blue total $2s$ and red total $2s$.

Alternative Approaches

A more synthetic proof of the first part uses directed angles. For each side of the square, the two radii to its intersection points with the circle are symmetric with respect to the perpendicular from the center to that side. The sum of the corresponding central angles is therefore fixed. Pairing opposite sides and adding these relations yields the equality of the total blue and red central angles without introducing coordinates.

Another approach to the second part is to regard the four perimeters together. The straight portions of all four curvilinear triangles cover each side of the square exactly once. Since opposite colors alternate around the square, each color receives exactly half of the total side contribution, namely $2s$. Combined with the arc equality, the perimeter equality follows immediately. The coordinate argument above is preferable because it gives a direct and completely explicit derivation of the arc equality.